Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have an exercise in which I need to determine which functions are continous for all points

Note: $\lfloor x\rfloor$ is floor of $x$.

Function I need to consider is $f(x)= \lfloor x\rfloor+\sqrt{x- \lfloor x\rfloor}$

How can I prove that this function is continuous at integer values if neither $\lfloor x\rfloor$ nor $\sqrt{x- \lfloor x\rfloor}$ are continuous at integer values?

I can determine it visually from graph and intuitively when considering values close to integers from both sides, but how can I rigorously prove it?

share|improve this question

3 Answers 3

Hint: For any $n\in \mathbb{N}$, we have $$\lim_{x\to n-} \lfloor x\rfloor=n-1$$ and $$\lim_{x\to n+} \lfloor x\rfloor=n$$

share|improve this answer

It's simple to see that it suffices to prove the continuity of $f$ on the integer. So let $n\in\Bbb Z$ then:

$$\lim_{x\to n^+}f(x)=\lim_{x\to n^+}n+\sqrt{x-n}=n$$and$$\lim_{x\to n^-}f(x)=\lim_{x\to n^-}n-1+\sqrt{x-n+1}=n-1+1=n$$ so $$\lim_{x\to n^+}f(x)=\lim_{x\to n^-}f(x)=n=f(n)$$ hence $f$ is continuous at $n$ and we conclude.

share|improve this answer

$1-{\bf 1}_{\Bbb Q}$ and ${\bf 1}_{\Bbb Q}$ are everyhere discontinuous, but their sum is $1$, which is continuous.

share|improve this answer
6  
This isn't really an answer. –  vadim123 Mar 16 at 20:11
1  
@vadim123 I think it is quite relevant. –  Pedro Tamaroff Mar 16 at 20:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.